The solubility product of PbS at `25^@ C `is `9 xx 10^(-28)`. What is its solubility in pure water and in `2 xx 10^(-12) M Na_2 S` solution ?

Explanation

Solubility of lead sulphide is given as, `PbS_((s)) hArr Pb_((aq))^(2+)+ S_((aq))^(2-)`
Solubility product`, Kp_(sp) =9 xx 10^(-28)`
Solubility of pbs in pure water `= sqrt(K_(sp)) = 3 xx 10^(-14) mol L^(-1)`
Solubility of pbs in `2 xx 10^(-2) M Na_a S` solution
`=(K_(sp))/([S^(2-)]) = (9 xx 10^(-28) )/( 2 xx 10^(-12)) = 4.5 xx 10^(-16) L^(-1)`